I have a question I haven't been able to find a direct answer to that I presume is true but I am unable to show. We know these two following results on the mellin transform.
If $$\int_0^\infty |f(x)|x^{\sigma - 1}\,dx < \infty$$ for all $a < \sigma < b$ Then
$$F(s) = \int_0^\infty f(x)x^{s-1}\,dx$$ is holomorphic for $a < \Re(s) < b$ and furthermore:
$$f(x) = \frac{1}{2 \pi i} \int_{\sigma - i\infty}^{\sigma + i\infty}F(s)x^{-s}\,ds$$
And conversely, if $F$ is holomorphic on $a < \Re(s) < b$ and
$$\int_{\sigma - i \infty}^{\sigma + i \infty} |F(s)|\,ds<\infty$$
then if
$$f(x) = \frac{1}{2 \pi i} \int_{\sigma - i\infty}^{\sigma + i\infty}F(s)x^{-s}\,ds$$
we get
$$\int_0^\infty f(x)x^{s-1}\,dx = F(s)$$
My question is: Does: $$\int_{\sigma - i \infty}^{\sigma + i \infty} |F(s)|\,ds<\infty\,\,\Rightarrow \,\,\int_0^\infty |f(x)|x^{\sigma - 1}\,dx < \infty$$
I'm thinking yes because the mellin transform is a modified fourier transform and this is true for the fourier transform... Thanks any help would be greatly appreciated.